Gravity Main with Low Outlet Pressure
A rural water supply scheme proposes a 2.5 km gravity main from an overhead tank to a village. The tank water level is 18 m above the village outlet. The design demand is 12 L/s. The available pipe has an internal diameter of 150 mm and a Darcy friction factor of 0.025. Minor losses are estimated at 10% of the friction loss. The engineer's estimate assumes the scheme will work because the static head is 18 m. During commissioning, the outlet pressure is only about 1 m and the measured flow is much less than 12 L/s. You must decide whether the problem is primarily excessive head loss, air locking, a partially closed valve, or higher demand, and recommend the next steps. The pipeline has no air-release valves at the high points.
1. Compute the head loss and residual pressure at the design flow.
Read the worked answer
Q = 12 L/s = 0.012 m3/s. Pipe area A = pi/4 x 0.150^2 = 0.01767 m2. Velocity V = Q / A = 0.012 / 0.01767 = 0.679 m/s.
Friction loss hf = f L V^2 / (2 g D) = 0.025 x 2500 x 0.679^2 / (2 x 9.81 x 0.150) = 9.8 m.
Minor loss = 10% of 9.8 = 0.98 m. Total head loss = 9.8 + 0.98 = 10.8 m.
Residual head at the outlet at design flow = 18 - 10.8 = 7.2 m. So at 12 L/s the pipe should still give roughly 7 m of pressure head, not 1 m.
2. The measured outlet pressure is only about 1 m and flow is much lower than design. What is the likely cause and what checks would you order?
Read the worked answer
At a lower flow, friction loss should be lower, not higher, so the low pressure is not explained by the design head-loss calculation. The most likely causes are air locking at high points, a partially closed or jammed valve, an obstruction or debris in the pipe, a lower actual tank level than assumed, or a break/leak that is diverting flow.
Order these checks: verify the actual water level in the tank; open and inspect all valves; install or open air-release valves at high points; measure pressure at several points along the main; check for blockage by flow and pressure tests; and inspect for leaks. Do not conclude that the pipe diameter is inadequate until these field checks are done.
3. If the village demand rises to 20 L/s, will the existing 150 mm pipe work? Show the comparison and one alternative.
Read the worked answer
At 20 L/s, V = 0.020 / 0.01767 = 1.132 m/s. Head loss scales approximately with V^2, so total loss at 20 L/s is about 10.8 x (1.132 / 0.679)^2 = 10.8 x 2.78 = 30.0 m.
This is greater than the available 18 m static head, so the 150 mm pipe cannot deliver 20 L/s by gravity. Alternatives include increasing the diameter or raising the tank/adding a booster. For example, a 200 mm pipe gives V = 0.020 / 0.0314 = 0.637 m/s; friction loss is about 6.5 m and minor loss about 0.65 m, total about 7.1 m, leaving about 10.9 m residual head. A larger diameter is therefore a technically sound option, subject to cost and site verification.